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Question Number 183594 by greougoury555 last updated on 27/Dec/22

log _(0.5) (√(1+x)) + 3log _(0.25) (1−x)= log _(1/16) (1−x^2 )^2 +2

$$\:\:\:\:\:\mathrm{log}\:_{\mathrm{0}.\mathrm{5}} \:\sqrt{\mathrm{1}+{x}}\:+\:\mathrm{3log}\:_{\mathrm{0}.\mathrm{25}} \left(\mathrm{1}−{x}\right)=\:\mathrm{log}\:_{\mathrm{1}/\mathrm{16}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\: \\ $$

Answered by hmr last updated on 27/Dec/22

• log _c (a^b ) = b log_c (a) • log_c^b (a) = (1/b) log_c (a) log _(0.5) ((√(1+x))) + (3/2) log_( 0.5) (1−x) = (1/4) log _(0.5) (1−x^2 )^2 + 2 log _(0.5) ((√(1+x))) + log_( 0.5) (1−x)^(3/2) = log _(0.5) (1−x^2 )^(1/2) +2log _(0.5) (0.5) log _(0.5) ((√(1+x)))(1−x)^(3/2) = log _(0.5) (1−x^2 )^(1/2) + log _(0.5) ((1/4)) log _(0.5) ((√(1−x^2 )))(1−x) = log _(0.5) ((1/4))(1−x^2 )^(1/2) ((√(1−x^2 )))(1−x) = ((1/4))((√(1−x^2 ))) 1−x = (1/4) x = (3/4) If ((√(1−x^2 ))) = 0 ; 1−x^2 = 0 → x = ±1 but x = ±1 is not in domain.

$$\bullet\:\mathrm{log}\:_{\mathrm{c}} \left(\mathrm{a}^{\mathrm{b}} \right)\:=\:\mathrm{b}\:\mathrm{log}_{\mathrm{c}} \:\left(\mathrm{a}\right) \\ $$$$\bullet\:\mathrm{log}_{\mathrm{c}^{\mathrm{b}} } \:\left(\mathrm{a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{b}}\:\mathrm{log}_{\mathrm{c}} \:\left(\mathrm{a}\right) \\ $$$$ \\ $$$${log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\sqrt{\mathrm{1}+{x}}\right)\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:{log}_{\:\mathrm{0}.\mathrm{5}} \:\left(\mathrm{1}−{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$${log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\sqrt{\mathrm{1}+{x}}\right)\:+\:{log}_{\:\mathrm{0}.\mathrm{5}} \:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\mathrm{2}{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\mathrm{0}.\mathrm{5}\right) \\ $$$${log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\sqrt{\mathrm{1}+{x}}\right)\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\left(\mathrm{1}−{x}\right)\:=\:{log}\:_{\mathrm{0}.\mathrm{5}} \:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\left(\mathrm{1}−{x}\right)\:=\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$\mathrm{1}−{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$${If}\:\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:=\:\mathrm{0}\:;\: \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} \:=\:\mathrm{0}\:\rightarrow\:{x}\:=\:\pm\mathrm{1} \\ $$$${but}\:{x}\:=\:\pm\mathrm{1}\:{is}\:{not}\:{in}\:{domain}. \\ $$

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